What is the solution for y^{''}+4y^'+3y=4x^2+5 ?

The solution for y^{''}+4y^'+3y=4x^2+5 is y=c_{1}e^{-x}+c_{2}e^{-3x}+(4x^2)/3-(32x)/9+149/27

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y^{''}+4y^'+3y=4x^2+5

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 x^{2} + 5

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} - \frac{32 x}{9} + \frac{149}{27}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 4 x^{2} + 5

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} - \frac{32 x}{9} + \frac{149}{27}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} - \frac{32 x}{9} + \frac{149}{27}

What is the solution for y^{''}+4y^'+3y=4x^2+5 ?
The solution for y^{''}+4y^'+3y=4x^2+5 is y=c_{1}e^{-x}+c_{2}e^{-3x}+(4x^2)/3-(32x)/9+149/27